Question

Let $f\left(x\right)=\left\{\begin{array}{cc}\frac{x-4}{\left|x-4\right|}+a,& x<4\\ a+b,& x=4\\ \frac{x-4}{\left|x-4\right|}+b,& x>4\end{array}.\right.$
Then, f (x) is continuous at x = 4 when
(a) a = 0, b = 0
(b) a = 1, b = 1
(c) a = −1, b = 1
(d) a = 1, b = −1.

Answer 1

(d) a = 1, b = −1.

Given: $f\left(x\right)=\left\{\begin{array}{c}\frac{x-4}{\left|x-4\right|}+a,\mathrm{if}\mathrm{x}<4\\ a+b,\mathrm{if}\mathrm{x}=4\\ \frac{x-4}{\left|x-4\right|}+b,\mathrm{if}\mathrm{x}>4\end{array}\right.$

We have
(LHL at x = 4) = $\underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(4-h\right)$

$=\underset{h\to 0}{\lim }\left(\frac{4-h-4}{\left|4-h-4\right|}+a\right)=\underset{h\to 0}{\lim }\left(\frac{-h}{\left|-h\right|}+a\right)=a-1$


(RHL at x = 4) = $\underset{x\to 4^{+}}{\lim }f\left(x\right)=\underset{h\to 0}{\lim }f\left(4+h\right)$

$=\underset{h\to 0}{\lim }\left(\frac{4+h-4}{\left|4+h-4\right|}+b\right)=\underset{h\to 0}{\lim }\left(\frac{h}{\left|h\right|}+b\right)=b+1$

Also,
$f\left(4\right)=a+b$

If f(x) is continuous at x = 4, then

$\underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{x\to 4^{+}}{\lim }f\left(x\right)=f\left(4\right)$

$\Rightarrow a-1=b+1=a+b$
$\Rightarrow a-1=a+b\mathrm{and}b+1=a+b$
$\Rightarrow b=-1\mathrm{and}a=1$